Consider a specific reaction with ΔH° = 66.8 kJ/mol and ΔS° = 105.3 J/mol•K. Assuming that these values do not change with temperature, at what temperature range is the reaction spontaneous in the forward direction?The reaction is spontaneous for temperatures (less than, greater than) T = ________ K.
Question
Consider a specific reaction with ΔH° = 66.8 kJ/mol and ΔS° = 105.3 J/mol•K. Assuming that these values do not change with temperature, at what temperature range is the reaction spontaneous in the forward direction?The reaction is spontaneous for temperatures (less than, greater than) T = ________ K.
Solution
The spontaneity of a reaction can be determined by the sign of the Gibbs free energy change (ΔG°). If ΔG° is negative, the reaction is spontaneous. The relationship between ΔG°, ΔH°, and ΔS° is given by the equation:
ΔG° = ΔH° - TΔS°
We are asked to find the temperature (T) at which the reaction becomes spontaneous, i.e., when ΔG° becomes negative. This occurs when ΔH° - TΔS° < 0.
Rearranging the equation gives:
T > ΔH° / ΔS°
Substituting the given values:
T > 66.8 kJ/mol / 105.3 J/mol•K
Note that we need to convert ΔH° from kJ/mol to J/mol to match the units of ΔS°.
So, T > 66800 J/mol / 105.3 J/mol•K = 634.38 K
Therefore, the reaction is spontaneous for temperatures greater than T = 634.38 K.
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